Problem: Evaluate the Maclaurin series. $1+\ln28+\dfrac{{{(\ln 28)}^{2}}}{2!}+\dfrac{{{(\ln28)}^{3}}}{3!}+\ldots +\dfrac{{{(\ln 28)}^{n}}}{n!}+\ldots $ Choose 1 answer: Choose 1 answer: (Choice A) A $28$ (Choice B) B $\ln(28)$ (Choice C) C $e^{28}$ (Choice D) D $\cos(28)$
Answer: Note that the given series has terms that do not alternate in sign that have all possible factorials in the denominator. This suggests that we are dealing with something about $~e^x$. Recall the Maclaurin series for $~e^x$. $ e^x=1\text{ }+\text{ }x\text{ }+\text{ }\frac{{{x}^{2}}}{2!}\text{ }+\text{ }\frac{{{x}^{3}}}{3!}\text{ }+\ldots +\text{ }\frac{{{x}^{n}}}{n!}\text{ }+\ldots $ Hence, we see that the given series converges to $~e^{\ln 28}$. $e^{\ln 28}=28$